How can I print a list with my own type of elements in OCaml?

Question

Here is the type of elements:

type listIS=Vide | I of int*listIS |S of string*listIS ;;

let l=I(5,I(3,S("3",I(0,S("",Vide)))));;

And my variant :

let rec afficheIS l=match l with
  |Vide ->""
  |I (i,l) -> (string_of_int i)^(afficheIS l)
  |S (s,l) -> (s)^(afficheIS l);;

But how can I change to give me the result :

5 3 3 0
-:unit = ()

Answer 1

I've several suggestions, on how to improve your code.

First of all is the beauty, it matters for those who read the code (not for the compiler). Please, consider using spaces between lexemes. It will be much easier to read your code.

Next, there is no need to put this ;;

And then I'd like to suggest a more cleaner implementation of the printing function

open Printf

type listIS =
  | Vide
  | I of int * listIS
  | S of string * listIS


let rec fprintIS out l = match l with
  | Vide    -> fprintf out "\n"
  | I (i,l) -> fprintf out "%d %a" i fprintIS l
  | S (s,l) -> fprintf out "%s %a" s fprintIS l


let l=I(5,I(3,S("3",I(0,S("",Vide)))))

let printIS = fprintIS stdout
let () = printIS l

I hope that you're familiar with printf function. If not, then I can try to find out some other solution, that will be easier for you to understand.

In this solution (I've prettified the code, as you may noticed) I use "%a" format specifier. It requires two arguments: first is the function that prints a value of an arbitrary type and the second is the value of that type. Using specifier you can build combine your printers.

Answer 2

Your problem is here: let rec afficheIS l=function your function afficheIS expect an argument l (that is never used) and you call afficheIS recursively you don't provide the l argument. Actually this l corresponds to nothing so you should remove it.

Answer 3

Solved:

 let rec afficheIS =function
      |Vide ->print_string "\n"
      |I (i,l) ->(print_string (string_of_int i));(afficheIS l);
      |S (s,l) -> print_string(s);(afficheIS l);;